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# If n is a positive integer...

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If n is a positive integer... [#permalink]  29 Jun 2006, 02:26
What is the best way to solve this question I mean what number properties/formula can be used in addition to brute force to solve it.
Thanks for your help.
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If n is a positive integer and the product of all the integers from 1 to n, inclusive, is divisible by 990, what is the least possible value of n?

A. 8

B. 9

C. 10

D. 11

E. 12
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[#permalink]  29 Jun 2006, 04:58
D

990 = 9*10*11

11 is the largest number hence n should be atleast 11
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Re: If n is a positive integer... [#permalink]  29 Jun 2006, 07:37
ishtmeet wrote:
What is the best way to solve this question I mean what number properties/formula can be used in addition to brute force to solve it.

Let product of numbers is X. Then

X/990 = A, where A is an integer.

For A to be an integer, all the factors of 990 must be present in X.

Factors of 990 = 9 * 11 * 5 * 2.

This means 11 must be there in X. Thats why answer is 11.
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SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

Re: If n is a positive integer...   [#permalink] 29 Jun 2006, 07:37
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